While reading the solutions of an exercise, I found the following statement:
Where $d$ is some operator and $I$ is the identity operator from $L^2$ to $L^2$. I managed to get the formula for $(I-dd^*)g$ that is written below, however, I couldn't draw the same conclusion. I checked in the lecture notes and I couldn't find any reference regarding the relationship between Fourier coefficients of a function in the image of an operator, and the operator being compact. If it helps, It is pretty easy to show that this operator is bounded.
Any ideas?
Let $\mathcal{H}$ be a Hilbert space with orthonormal basis $\{ e_k \}_{k=-\infty}^{\infty}$. Your operator is the same as the diagonal operator $D$ given by $$ Df = \sum_{k=-\infty}^{\infty}\frac{1}{1+k^2}\langle f,e_k\rangle e_k. $$
To show that $D$ is compact, let $\{ f_n \}_{n=1}^{\infty}$ be bounded sequence in $\mathcal{H}$ with bound $M$; it must be shown that $\{ Df_n \}_{n=1}^{\infty}$ has a convergent subsequence.
Every coordinate sequence $\{ \langle f_n,e_k\rangle\}_{n=1}^{\infty}$ is bounded by $M$; this bound holds for all $k$. Using Cantor diagonalization, there is a subsequence $\{ f_{n_m} \}_{m=1}^{\infty}$ such that $\lim_{m}\langle f_{n_m},e_k\rangle$ converges for every $k$ to some limit $\alpha_{k}$. The sequence $\{\alpha_k\}$ is uniformly bounded by $M$; therefore $y=\sum_{k=-\infty}^{\infty}\frac{\alpha_k}{1+k^2} e_k \in \mathcal{H}$. And, $$ \left\|Df_{n_m}-y\right\|^2 = \sum_{k=-\infty}^{\infty}\frac{1}{1+k^2}|\langle f_{n_m},e_k\rangle-\alpha_k|^2 $$ converges to $0$ as $m\rightarrow\infty$. (You can apply the Lebesgue dominated convergence Theorem applied to a discrete measure, or you can argue directly.) Hence, $$ \lim_{m\rightarrow\infty}Df_{n_m}=\sum_{k=-\infty}^{\infty}\frac{1}{1+k^2}\alpha_k e_k $$ So $D$ is compact.